Question

2. A rectangular overhanging beam and loads are shown in the following figure. A. Draw the shear and bending-moment diagrams B. Determine the maximum absolute value of the shear and absolute maximum bending stress. 16KN/m 45KN (15pts) Sem 4m 2m- 8 cm

Answer 1

We have to start with finding the support reactions. Then we can write the shear force and bending moment equations for every section of the beam and can make the diagrams also. From diagrams we can find the maximum values of shear force and maximum positive bending moment to find the stresses. Please find the attached solution.

Robi 45 KN 16 kN/m B 2 m. 4m. RA Rc {fy to RAT Rc 16x4 + 45 = 109 -0 EMA EO Rex 4 45X6 + 16X4 X2 Re= 99.5kN RA = 9.5 KN Shear force. Between Be » 45 KN (constant). Between ca Fx = 45- 99.5+ 16(x-2) fx= 162- 86.5 Exto at x = 5.4 m E(Point). from Right at a=6m. fa= 96-8615 9,5 km.

SED 45KN. 9.5 KN. + D B 54.5 KN. Bending Moment s ved True → Between Be. Mal = - 452 Between CAO Ma -45% + 99.5 (2-2) 16(x-2)2 2 → 54.5x - 199 - 8(2-2)2 putting Maco < d=6m. (Point A] x=4.81 m. (point ou at B. 2= 2 m. MB = 90 KN.M ALE. x=5.4 ME 54.5*5.4 – 199 - 0 (54-212 ME= 2.82 kNom

B.M.D. 4.8m. ik 2.82 knoop P-3 + A 13 E 1 90kn-m. b maximum shear force Fmax. = -54.5 KN (At point & & marm bending moment Mmaxi 90 kNam (at point a >> + 2.82 KN-m Cat Ev Area of beam (xaseen) A = bxh h = 0.08m. oodX0.05 b=0.05m marm shear stress max. = Fmax. 13625 kN/m2 A Emax. = 13.625 MPa

Area MOI. I. bh3 0.05 % (0.00) 12 12 mexia o od 2 0.04 mi % B.M. 2.82 KNom. Махт positive so from bending ecreation. 11 G1s Elt 2.82 x 0.04 mani 0.05 x 0.0819/12 max- = 52.9 MPa